Infinite sum compression using the inverse of a formal power series

• Infinite sum compression using the inverse of [formal power series

• [Polynomial and Formal Power Series (2) Derivation of Solutions by Expression Transformation | maspy's HP https://maspypy.com/%e5%a4%9a%e9%a0%85%e5%bc%8f%e3%83%bb%e5%bd%a2%e5%bc%8f%e7%9a%84%e3%81%b9%e3 I have paraphrased a passage from %81%8d%e7%b4%9a%e6%95%b0%ef%bc%88%ef%bc%92%ef%bc%89%e5%bc%8f%e5%a4%89%e5%bd%a2%e3%81%ab%e3%82%88%e3%82%8b%e8%a7%a3%e6%b3%95 to the point where I can to a point where I can understand it.

• inverse element

  • $A = \sum_{n=0}^\infty x^n$
  • $B = 1-x$
  • At this time [$ A \times B = 1
    • $A = 1 + x + x^2 + \ldots$ (1)
    • $xA = x + x^2 + \ldots$ (2)
    • $(1-x)A = 1$ (1) - (2)
    • Common techniques for calculating sums of sequences of equal ratios
  • Generally when F is a formal power series without a constant term such that $[x^0]F = 0$,
    • $(1-F) \times (\sum_{n=0}^\infty F^n) = 1$
    • The proof requires defining a phase in the formal power ring and defining convergence

Count up the number of ways to express N as a sum of positive integers. However, distinguish between different orders of sums.

• $F = \sum_{n=1}^\infty x^n$,[$ G = \sum_{n=0}^\infty F^n
• $[x^n]G$ is the answer
• Replace the infinite sum of G with the inverse original
  • $G = \sum_{n=0}^\infty F^n = \frac{1}{1 - F}$
• The infinite sum of F is also replaced by the inverse element
  • $F = \sum_{n=1}^\infty x^n = \sum_{n=0}^\infty x^{n+1} = x\sum_{n=0}^\infty x^n = \frac{x}{1-x}$
• Substitute F for G
  • $G = \frac{1}{1 - F} = \frac{1}{1-\frac{x}{1-x}} = (1-x) \frac{1}{1-2x}$
• where the $2x$ part is a formal power series, so we can return to infinite sums, which we will call J
• $J = \frac{1}{1-2x} = \sum_{n=0}^\infty (2x)^n$
  • $[x^n]J = 2^n$
  • From here $[x^n] G = x^nJ = [x^n]J - [x^{n-1}]J = 2^n - 2^{n-1} = 2^{n-1}$

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